Oscillations 2008 prelim_solutions

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Oscillations 2008 prelim_solutions

  1. 1. Oscillations 2008 Prelim Solutions 1. B 2. C 3. C 4. a. i. Upwards ii. Replace the volume of object submerged with a liquid column of height d and uniform cross-sectional area A. Weight of liquid above = Adρg Pressure on bottom due to liquid = W/A = dρg iii. Pressure due to liquid P = pressure difference betw top and bottom surface (need to be mentioned somewhere) Hence Force upwards = PA = Vρg Where V is the vol displaced. b. i. When pushed down a further distance x, New force upwards = A(d+x) ρg But initially floating, hence Adρg = weight downwards Hence resultant force = Axρg c. i. Since the resultant force is directly proportional to x and it acts in the opp direction to the displacement, the resultant motion is in SHM as it follows the defining eqn of SHM. ii. comparing iii. As the block oscillates in the liquid, it has to overcome the resistive (viscous) forces due to the liquid, hence energy of the oscillating system decreases, resulting in damped oscillations. iv. Correct shape - Decreasing amplitude shown (Periods not impt) Shown for 3 periods correctly d. i. Forced Oscillation: The water waves acts as an external periodic driving force continuously supplies energy to the damped system to compensate for the loss of
  2. 2. energy due to the loss of energy which occurred during the oscillation of the system. The system will oscillate with constant amplitude and with the frequency of the water waves. ii. Resonance: Phenomenon where there is maximum transfer of energy from the forced oscillator such that the system oscillates with the maximum amplitude. It occurs when at the point when the frequency of the water waves called the driving frequency is close to the natural frequency of the system. Note: If answer is general statement – not related to question - max 3 marks 1. f = v/λ = 3 Hz 2. 0.0788 kg (allow ecf) 3. When the block has absorbed some water, the natural frequency of the system decreases, system no longer at resonance, hence the amplitude of oscillation will be lower. 5. A 6. B 7. (a) – (c) are not SHM related d. i. (x and a are in opposite directions) Hence, 1. the period of the oscillation = 3.14 s 2. the kinetic energy as it passes through the equilibrium position = 1m(Aω) 2 = (5.0)(0.200x2) = 0.400 J 3. the maximum net force experienced = mAω 2 = 5.0 x 0.200 x 22 = 4.00 N
  3. 3. iii. iv. x = 0.2 cos ωt = 0.2 cos (2 x 3) = 0.192 m Distance from expected position = 0.2 – 0.192 = 0.008 m 8. B The phase difference is not constant as it is shown that at time, the two waves are in phase and after a while the two waves are anti-phase. 9. C Maximum displacement = 5 That occurs when sin (20π t) = 1 20π t = π/2 t = 1/40 s Half way through equilibrium position = 2.5 m, that occurs when sin (20π t) = ½ 20π t = 5π/6 (It has to be after π/2) t = 1/24 s Hence time taken = 1/24 – 1/40 = 1/60 s that is between 1/80 and 1/40 s 10. a. i. initial P.E. = (1000)(0.1) 2 = 5.00 J ii.
  4. 4. b. Initially, the amplitude of oscillations decreases to zero, then it increases and reaches a constant steady value. c. (circular motion related) 11. D Since x = 0 m at t = 0 s, The x-t graph must take a sine function. x = xo sin ωt or x = - xo sin ωt Hence, a = - ω 2 xo sin ωt or a = ω 2 xo sin ωt 12. a. The force acting on the ball is not proportional to its displacement. [B1] b. i. [B1] ii. [M1] Comparing with the equation obtained with iii. The mass of the spring is not taken into consideration. [B1] c. Circular motion related question 13. D
  5. 5. Using x = 2 cos ωt, at t = 1.5 s x = 2 cos (π x 1.5) = 0 The displacement is zero. Since the period is 2.0 s and 1.5 s is of the period, thus the distance travelled by particle is 6.0 cm 14. D Since total energy ∝ amplitude of oscillation (ET = ½ mω 2 x0 2 ) the total energy decays exponentially. 15. a. a = -ω 2 x [1] b. i. Angular frequency ω = 2πf = 2π ( = 17.6 rad s -1 [1] ii. x = 0.04 cos 17.6 t iii. Maximum speed v = ωx0 = 17.6 x 0.04 [1] = 0.704 m s -1 [1] iv. [1] shape [1] indicate 0.04 m AND 0.704 m s -1 v. [1] decreasing amplitude [1] period remain same or increase 0.704 m s-1
  6. 6. 16. D A In steady state the system vibrates at the frequency of the driving force. Under the influence of a driving force, the system will oscillate at the driving frequency. This is known as forced oscillations. B The amplitude of vibration becomes very large when the frequency of the driving force is close to the natural frequency of vibration of the system. When the driving frequency is close to the natural frequency, the system vibrates with maximum amplitude. This phenomenon is called resonance. C The amplitude of vibration remains finite if damping forces are present. Effects of damping on amplitude of oscillations are shown in the graph on the right. D At resonance the displacement of the system is in phase with the driving force. This question can only be answered by process of elimination. Options A to C are taught to you as facts, therefore the only answer to the question has to be D. 17. A Acceleration of the bouncing ball is not directly proportional to displacement; moreover, there is no equilibrium point. For all other options, acceleration is directed towards the equilibrium point, and value changes with displacement. (For SHM, a = -ω 2 x)
  7. 7. 18. a. a = - ω 2 x [1] ω = = 8.72 rad s -1 [1] b. Equation for motion: x = - xo cos ωt [1] (Since starting point is at lowest point of motion) c. Resonance occurs. The frequency of passing over the bumps matches the natural frequency of the truck. OR the time between the bumps matches the natural period of the truck. [1] 19. D 20. a. Force acting on the base of the tube = upthrust = ρlAg b. i. ii. When the test-tube is displaced downwards by x, U’ = since at equilibrium, ρlAg = mg  Resultant force is upwards  Acceleration is directed upwards (opposite to the direction of displacement) Consider the free body diagram of the test-tube when it is displaced downwards by x, Since ρlAg = mg, we obtain ρxAg = ma, where a α x (acceleration is directly proportional to the magnitude of the displacement)
  8. 8. iii. Considering SHM, mω 2 x = ρxAg mω 2 = ρAg But, ρlAg = mg giving ρAg = Hence, 21. D Equation is in the form x = xo sin (ωt) vmax = ωxo = (6π)(5.0 x 10 -3 ) = 9.4 x 10 -2 m s -1 22. A Z is the driver causing oscillations of X and Y. By varying the length of Z, the driver frequency is being changed. Since X is made of polystyrene, it is subjected to air resistance, so its motion is more damped, thus the curve is less sharp. 23. A For SHM, a α – x As a = - 10x, therefore ω 2 = 10 2πf = 24. B 25. C 26. a. Simple harmonic motion is defined as the motion of an object whose acceleration a, is directly proportional to its displacement x from a fixed point (equilibrium position) and is always directed towards that fixed point. The defining equation of simple harmonic motion is a = - ω 2 x. The minus sign indicates that a is always in the opposite direction to x. b. When the object is moving towards the equilibrium position, its acceleration and velocity are in the same direction. When the object is moving away from the equilibrium position, its acceleration and velocity are in opposite direction. c. 27. B
  9. 9. 28. a. An oscillatory motion in which the acceleration (or restoring force) is always proportional to and opposite in direction to the displacement from the equilibrium position. [1] b. F = m a = m ω 2 x [1] = (0.8) (2πf) (2 x 10 -2 ){f = 2 Hz} = 2.53 N [1] Direction = downwards [1] c. KE = mv 2 = [1] = [1] = 0.011 J d. Loss in GPE + Loss in KE = Gain in EPE {2 marks for three, 1 mark for 2} 29. a. i. By Newton’s 2nd Law, - k x = m a [1] a = - k x / m [1] ii. Equation shows acceleration is directly proportional to negative of displacement [1] since k and m are constant. So motion is SHM. [1] b. E = ½ k x 2 = ½ (2.8 x 106) (0.70) 2 = 690 kJ [1] c. E = ½ m ω 2 A 2 2 2 '' A A E E  A’ = 0.49 m [1] 30. B 31. C 32. A 33. a. a = −(2πf) 2 x b. Eo = (1.0 × 10 −12 )(3.00 × 10 8 ) = 3.0 × 10 −4 V m −1 c. ao = (1.60 × 10 −19 )( 3.0 × 10 −4 ) / ( 9.11 × 10 −31 ) = 5.269 × 10 7  xo = 5. 269 × 10 7 ÷ (2 × 200 × 10 3 ) 2 = 3.34 × 10 −5 m d. ∆x∆p ≥ h/(4π)
  10. 10.  ∆v ≥ h {(4π)(3.34 × 10 −5 )(9.11 × 10 −31 )} = 1.73 m s −1 = 2 m s −1 (1.s.f) 34. C The kinetic energy will be more than the potential energy at this point. 35. B 36. a. b. x = x0 cos [(2πf)t] = (1.25 x 10 -2 ) cos [(2π x 3.0) (0.02)] = 0.012 m Displacement required = 0.0125 – 0.012 = 5.0 x 10 -4 m c. i. The surface water waves act as the external driving oscillator to the oscillating block in this case. When the frequency of the water waves matches the natural frequency of the block, resonance occurs and this results in a large increase amplitude in the oscillation. ii. The block no longer resonates and amplitude is lower. d. i & ii

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